EXPLANATION OF LEAD IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 22 , 2015 ' ' ' INTRODUCTION' Lead is a chemical element with symbol Pb and atomic number 82. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Lead including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25p6 4f 145d106s2 6px16py1 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of lead (from E1 to E5 ) are the following: E1 = 7.42 , E2 = 15 , E3 = 31.9 , E4 = 42.32 , and E5 = 68.8 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my new paper of 2008. ' ' EXPLANATION OF E1 = 7.42 eV = -E(6px1) The electron charges (-80e) of the 80 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f 145d106s2) screen the nuclear charge (+82e) and for a perfect screening the one electron of 6p would provide an effective Zeff = ζ = 2. Here the two electrons of parallel spin repel from symmetrical positions the electron clouds of 6s2 and the 5d10 and lead to the deformation of electron clouds. Thus ζ > 2. Note that the E(6px1) represents the binding of 6px1 given by applying the Bohr formula as ' '''E1 = 7.42 eV = (6px1) = (-13.6057)ζ2/n2 Then using n = 6 we get ζ = 4.43 > 2. ' ' '''EXPLANATION OF E2 = 15 eV = -E(py1)' The E(py1) represents the binding energy of py1 given by applying the Bohr formula as E2 = 15 eV = -E(py1) = -(-13.6057)ζ2/n2 Then using n = 6 we get ζ = 6.3 > 2 . Here the ζ = 6.3 > 4.43 > 2 means that after the first ionization the one electron of 6p breaks more the symmetry and leads to a greater deformation of electron clouds. ' ' EXPLANATION OF E3 = 31.9 eV = -E(6s2) + E(6s1) According to the experiments the electrons of 6s2, 5d10, and 4f14 belong to the same energy level under n = 6. However the electron charges (-78e) of the 78 electrons of the following configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f145d10) screen the nuclear charge (+82e) and for a perfect screening the electrons of 6s2 would provide an effective ζ = 4. But the above electrons of 6s penetrate the 5d10 and lead to the deformation of electron clouds. Thus ζ > 4. Here the E(6s2) represents the binding energy of 6s2, while the E(6s1) represents the binding energy of 6s1, which appears after the first ionization of 6s2 . Note that the 6s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6s1 consists of one electron, we apply the Bohr formula as E(6s1) = (-13.6057)ζ2/n2 Therefore E3 = 31.9 eV = -E(6s2) + E(6s1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1144.3 = 0 and solving for ζ we get ζ = 9.82 > 4. Of course the two electrons of opposite spin (6s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a net mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 . EXPLANATION OF E4 = 42.32 eV = -E(6s1) As in the case of E3 the charges (-78e) of 78 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p64f145d10) screen the nuclear charge (+82e) and for a perfect screening we would have the same effective ζ = 4. However the one electron of 6s1 penetrates the 5d10 and leads to the deformation of electron clouds. Therefore we have the same ζ > 4. Here the E(6s1) represents the binding energy of (6s1) given by applying the Bohr formula as E4 = 42.32 eV = -Ε(6s1) = - ( -13.6057)ζ2 /62 Then solving for ζ we get ζ = 10.58 > 4 . Here the ζ = 10.58 > 9.82 > 4 means that after the ionization the one electron of 6s breaks more the symmetry and leads to a greater deformation of electron clouds. EXPLANATION OF E5 = 68.8 eV = -E(5d2) + E(5d1) It is of interest to note that the 5d10 consists of five pairs (10 electrons of 5d2, 5d2, 5d2, 5d2 , and 5d2 with opposite spin). Here the E(5d2) represents the binding energy of the first 5d2, while the E(5d1) represents the binding energy of 5d1, which appears after the first ionization of the first 5d2 . Here the charges (-68e) of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p64f 14) screen the nuclear charge (+82e ) and for a perfect screening we would have an effective ζ = 14. Note that the first 5d2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5d2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5d1 consists of one electron, we apply the Bohr formula as E(5d1) = (-13.6057)ζ2/n2 Therefore E5 = 68.8 eV = -E(5d2) + E(5d1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 2472.7 = 0 and solving for ζ we get ζ = 14.12 , which is nearly equal to the perfect screening with ζ = 14 . Category:Fundamental physics concepts